Question: In Problems , solve the equation.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}{\left(x+y+1\right)}^2\mathbf{-}{\left(x+y-1\right)}^2$

Definition of Initial Value Problem: By an initial value problem for an nth-order differential equation$\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{,}\mathrm{...}\mathbf{,}\frac{{\mathbf{d}}^n\mathbf{y}}{{\mathbf{dx}}^n}\right)\mathbf{=}\mathbf{0}\mathbf{,}$ we mean: Find a solution to the differential equation on an interval I that satisfies at x₀ the n initial conditions

$\begin{array}{c}\mathbf{y}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_0\mathbf{,}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_1\mathbf{,}\\ .\\ .\\ .\\ \frac{{\mathbf{d}}^{n-1}\mathbf{y}}{{\mathbf{dx}}^{n-1}}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_{n-1}\mathbf{,}\end{array}$

Where${\mathbf{x}}_0\in \mathbf{I}$ and${\mathbf{y}}_0\mathbf{,}{\mathbf{y}}_1\mathbf{,}\mathrm{...}\mathbf{,}{\mathbf{y}}_{n-1}$ are given constants.

Formulae to be used:

• Integration by parts: $\int \mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}\int \mathbf{vdu}$.
• ${\mathbf{a}}^2\mathbf{-}{\mathbf{b}}^2\mathbf{=}\left(a-b\right)\left(a+b\right)$
  
• $\int {\mathbf{e}}^{\mathrm{ax}}\mathbf{dx}\mathbf{=}\frac{{\mathbf{e}}^{\mathrm{ax}}}{a}\mathbf{+}\mathbf{C}$.

Given that,$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}{\left(x+y+1\right)}^2\mathbf{-}{\left(x+y-1\right)}^2\mathrm{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Evaluate the equation (1).

$\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}{\left(x+y+1\right)}^2\mathbf{-}{\left(x+y-1\right)}^2\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\left(\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{1}\mathbf{-}\left(x+y-1\right)\right)\left(\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{1}\mathbf{+}\left(x+y-1\right)\right)\\ \mathbf{=}\mathbf{2}\left(2x+2y\right)\\ \mathbf{=}\mathbf{4}\left(x+y\right)\end{array}$

Let us take $\mathbf{t}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}\to \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\mathbf{1}$.

Then,

$\begin{array}{c}\frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{4}\mathbf{t}\\ \frac{\mathrm{dt}}{\mathrm{dx}}\mathbf{=}\mathbf{4}\mathbf{t}\mathbf{+}\mathbf{1}\\ \frac{\mathrm{dt}}{4t+1}\mathbf{=}\mathbf{dx}\end{array}$

Now integrate the equation (2) on both sides.

$\begin{array}{c}\int \frac{\mathrm{dt}}{4t+1}\mathbf{=}\int \mathbf{dx}\\ \frac{\mathbf{ln}\left|4t+1\right|}{4}\mathbf{=}\mathbf{x}\mathbf{+}{\mathbf{C}}_1\\ \mathbf{ln}\left|4t+1\right|\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{C}\end{array}$

Simplify it.

$\begin{array}{c}\mathbf{4}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{=}{\mathbf{Ce}}^{4x}\\ \mathbf{4}\mathbf{t}\mathbf{=}{\mathbf{Ce}}^{4x}\mathbf{-}\mathbf{1}\\ \mathbf{t}\mathbf{=}\frac{{\mathbf{Ce}}^{4x}\mathbf{-}\mathbf{1}}{4}\end{array}$

Substitute the value of t.

$\begin{array}{c}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\frac{{\mathbf{Ce}}^{4x}\mathbf{-}\mathbf{1}}{4}\\ \mathbf{y}\mathbf{=}\frac{{\mathbf{Ce}}^{4x}\mathbf{-}\mathbf{1}}{4}\mathbf{-}\mathbf{x}\\ \mathbf{y}\mathbf{=}\frac{{\mathbf{Ce}}^{4x}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}}{4}\end{array}$

Hence, the solution of the given initial value problem is $\mathbf{y}\mathbf{=}\frac{{\mathbf{Ce}}^{4x}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{1}}{4}$.